Approximately 15 m/s is the speed of the car.
<u>Explanation:</u>
<u>Given:</u>
speed of sound - 343 m/s
You detect a frequency that is 0.959 times as small as the frequency emitted by the car when it is stationary. So, it can be written as,
If there is relative movement between an observer and source, the frequency heard by an observer differs from the actual frequency of the source. This changed frequency is called the apparent frequency. This variation in frequency of sound wave due to motion is called the Doppler shift (Doppler effect). In general,
Where,
- Observed frequency
f – Actual frequency
v – Velocity of sound waves
– Velocity of observer
- velocity of source
When source moves away from an observer at rest (
), the equation would be
By substituting the known values, we get
Approximately 15 m/s is the speed of the car.
Mass of first car = Initial mass (Mi) = 2 kg
Initial velocity (Vi) = 2 m/s
Mass of both cars together = Final mass (Mf) = 2 + 3 kg = 5 kg
Final Velocity (Vf) = ?
Applying law of conservation of momentum,
Mi x Vi = Mf x Vf
2 x 2 = 5 x Vf
Vf = 4/5 = 0.8 m/s
As the greater force of tension (by 81N) is exerted by the team on the right the rope will move to the right.
Answer:
u = 11.6 m/s
Explanation:
The end of a launch ramp is directed 63° above the horizontal. A skier attains a height of 10.9 m above the end of the ramp.
Maximum height, H = 10.9
Let v is the launch speed of the skier. The maximum height attained by the projectile is given by :
u = 11.6 m/s
So, the launch speed of the skier is 11.6 m/s. Hence, this is the required solution.
Answer:
Hello your question is poorly written below is the complete question
Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
answer :
a) 231.48 days
b) n = 3.125 * 10^15
Explanation:
Battery moved 10,000 coulombs
current rate = 0.5 mA
<u>A) Determine how long the clock run on the battery. use the relation below</u>
q = i * t ----- ( 1 )
q = charge , i = current , t = time
10000 = 0.5 * 10^-3 * t
hence t = 2 * 10^7 secs
hence the time = 231.48 days
<u>B) Determine how many electrons per second flowed </u>
q = n*e ------ ( 2 )
n = number of electrons
e = 1.6 * 10^-19
q = 0.5 * 10^-3 coulomb ( charge flowing per electron )
back to equation 2
n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )
hence : n = 3.125 * 10^15
